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The domain of a cartesian function from parametric equations

  • Thread starter nokia8650
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  • #1
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x = 2cot t

y = (sin t)^2

t is greater than 0 but less than or equal to pi/2


The cartesian can be found using trig identities to be:

y = 8/ (4+ x^2)


What would be the range of the cartesian equation? I think it would be x is greater than or equal to 0, since when t = pi/2, x = 0, and as t tends to 0, x tends to infinity.

Am I correct?

Thank you.
 

Answers and Replies

  • #2
HallsofIvy
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Your title says "domain" but in the body of your message you say "range". Which is it?

If you are thinking of the "cartesian equation" as y a function of x, then the domain is the set of all possible x values which is the set of all vaues of cot(t) for t between 0 and pi/2 and the range is the set of all y values which is the set of all values of (sin t)^2 for t between 0 and pi/2.
 
  • #3
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Sorry for the confusion! I meant the domain! Would my answer therefore be correct?

Thanks
 
  • #4
HallsofIvy
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Yes.
 

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